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This post has some answers that give some intuition as to the definition of the transpose. My rudimentary (perhaps inaccurate) understanding is that for a linear transformation $T: V \to W$, we're interested (why?) in a way to represent functionals on transformed points, $f \in W^*$, as functionals on the original points $T^\top(f) \in V^*$. Friedberg, Insel, Spence write:

For a matrix of the form $A = [T]_{\beta\to\gamma}$, the question arises as to whether or not there exists a linear transformation $U$ associated with $T$ in some natural way such that $U$ may be represented in some basis as $A^\top$

From where does "this question arise"? The notion of the existence of a transformation $T: V \to W$ does not obviously seem to imply the existence of a "dual" transformation from $W^*$ to $V^*$. Why do we want to "[represent] $U$ in some basis as $A^\top$" at all? And why do we care whether such a transformation exists? I would prefer an elementary explanation that doesn't invoke adjointness as a motivator (since it is not technically covered until later in this text), but this question of "why do we care about this" has been confusing me for a few days now. Thank you!

krt24
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    I find most books on linear algebra do an abysmal job motivating and explaining duality. You might find my video helpful. – blargoner Aug 03 '22 at 22:16
  • I kinda have no idea what they mean by that! – SBK Aug 03 '22 at 22:26
  • I disagree with the quote saying the question arises as to what linear transformation would have matrix $A^T$. I suppose it's technically true for a reader reading about the matrix $A^T$ wondering why it's a thing, but that is artificial. But I also think your claim that a linear map $T:V\to W$ does not obviously imply the existence of a linear map $W^\ast\to V^\ast$ as wrong: the map is precomposition-by-$T$, which shows up in things so it is a natural thing to think about. – anon Aug 03 '22 at 22:28
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    Consider column vectors $x$ and $y$ in some vector space $V$. Now given $x^Ty$ If I fix $x$ and allow $y$ to vary, we have a linear equation in the components of $y$. Now if I allow $x$ to vary we instead have linear functionals from $V$ to its base field. We're not really up to anything special here, it's just the symmetry of the equation $\sum x_k y_k$ that leads to this duel space since the $x$ and $y$ components can be swapped. Once that is in place general transposes follow. – CyclotomicField Aug 03 '22 at 22:37
  • @runway44 thank you, that is a good point. I suppose it's "natural" to consider the pullback of the composition $f \circ T$ to the function $T^\top(f) \in V^*$ -- but I feel like I'm still missing some context as to why the transformation from $f \circ T$ to its pullback $T^\top$ is so important to deserve this much attention at all! – krt24 Aug 03 '22 at 22:47
  • @krt24 They adjoint operator is useful when working in inner product spaces so the transpose naturally arises in that context. There are many applications for it Hilbert spaces which by proxy is physics. – CyclotomicField Aug 03 '22 at 23:25
  • @CyclotomicField true! I guess the reason I was trying to avoid the mention of adjointness was because I was trying to understand how the authors of this text were motivating the idea of the transpose (from the answers I'm seeing, it appears as though the answer is "they didn't do a good job"). But you bring up an excellent example of a slightly different interpretation that is (in my opinion) more straightforward. Also appreciate the illustration from your previous comment. – krt24 Aug 03 '22 at 23:47
  • Do Friedberg, Insel, Spence spend time looking for a natural linear map whose matrix the $A$ rotated 90º? – Mariano Suárez-Álvarez Aug 04 '22 at 06:51

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